Question

Suppose that while waiting for a big wave, a surfer is being propelled toward the beach in such a way that the distance (in meters) traveled is given by f (t) = 10t2 for 0 ≤ t ≤ 1, where t is in seconds. Find the velocity v 1 10 of the surfer at t = 1 10 sec.

Answer 1

The velocity at t=1/10s will be:

`v((1)/(10))=2(m)/(s)`

Step-by-step explanation:

The distance in meter as a function of time is:

`f (t) = 10t^2`, for 0 ≤ t ≤ 1 where t is in seconds.

We need to find the velocity
`v((1)/(10))` of the surfer at
`t=(1)/(10)s`.

For this we can use the definition of instantaneus velocity (in it's limit form):

`v(t) =  \lim_(\Delta t \to 0) (f(t+\Delta t)-f(t))/(\Delta t) =  \lim_(\Delta t \to 0) (10(t+\Delta t)^2 - 10t^2)/(\Delta t) \\=  \lim_(\Delta t \to 0) (10(t^2+2t\Delta t+\Delta t^2) - 10t^2)/(\Delta t) =  \lim_(\Delta t \to 0) (20t\Delta t+10\Delta t^2)/(\Delta t) \\=  \lim_(\Delta t \to 0) ((20t+10\Delta t)\Delta t)/(\Delta t) =  \lim_(\Delta t \to 0) 20t+10\Delta t = 20t`

At
`t=(1)/(10)s`, we have:

`v((1)/(10)) =  20(1)/(10)=2(m)/(s)`.